package graph;

/**
 * 给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
 * 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
 * 此外，你可以假设该网格的四条边均被水包围。
 * <p>
 * 示例 1：
 * 输入：grid = [
 * ["1","1","1","1","0"],
 * ["1","1","0","1","0"],
 * ["1","1","0","0","0"],
 * ["0","0","0","0","0"]
 * ]
 * 输出：1
 * <p>
 * 示例 2：
 * 输入：grid = [
 * ["1","1","0","0","0"],
 * ["1","1","0","0","0"],
 * ["0","0","1","0","0"],
 * ["0","0","0","1","1"]
 * ]
 * 输出：3
 *
 * @author Jisheng Huang
 * @version 20250516
 */
public class NumOfIslands_200 {
    /**
     * Loop through the given 2d char array. Do the dfs on the array
     *
     * @param grid the given 2d array
     * @return the number of isolated islands
     */
    public static int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int islands = 0;

        for (int i = 0; i < grid.length; ++i) {
            for (int j = 0; j < grid[0].length; ++j) {
                islands += dfs(grid, i, j);
            }
        }

        return islands;
    }

    /**
     * dfs traversal for the 2d array
     *
     * @param grid the given array
     * @param i    the x axis
     * @param j    the y axis
     * @return 1 if it is an isolated array, 0 otherwise
     */
    public static int dfs(char[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == '0') {
            return 0;
        }

        dfs(grid, i + 1, j);
        dfs(grid, i - 1, j);
        dfs(grid, i, j + 1);
        dfs(grid, i, j - 1);

        return 1;
    }

    public static void main(String[] args) {
        
    }
}
